let A_n = ∫_0 ^∞ e^(−nx^2 ) sin((x/n))dx with n integr not 0 1) calculate A_n 2) find lim_(n→+∞) A


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Question Number 37285 by math khazana by abdo last updated on 11/Jun/18

$l e t A_{n} = \int_{0}^{\infty} e^{- {n x}^{2}} s i n (\frac{x}{n}) d x w i t h n i n t e g r n o t 0$ $1) c a l c u l a t e A_{n}$ $2) f i n d {l i m}_{n \to + \infty} A_{n}$
Commented by math khazana by abdo last updated on 14/Jun/18

$A_{n} = \int_{0}^{+ \infty} e^{- {n x}^{2}} c o s (\frac{x}{n}) d x .$
Commented by prof Abdo imad last updated on 16/Jun/18
$changement (x/n)=t give A_n = ∫_0 ^∞ e^(−nn^2 t^2 ) cos(t) ndt =n ∫_0 ^∞ e^(−n^3 t^2 ) cost dt =(n/2) ∫_(−∞) ^(+∞) e^(−n^3 t^2 +it) dt (2/n) A_n = ∫_(−∞) ^(+∞) e^(−{(n(√n)t)^2 −it}) dt =∫_(−∞) ^(+∞) e^(−{ (n(√n)t)^2 −2 (i/(n(√n)))(n(√n)t) + ((i/(n(√n))))^2 −((i/(n(√n))))^2 }) dt = ∫_(−∞) ^(+∞) e^(−(n(√n) t −(i/(n(√n))))^2 −(1/n^3 )) dt =_(n(√n)t −(i/(n(√n))) = u) e^(−(1/n^3 )) ∫_(−∞) ^(+∞) e^(−u^2 ) du = (√π) e^(−(1/(n^3 ))) ⇒ A_n =((n(√π))/2) e^(−(1/(n^3 ))) .$
$c h a n g e m e n t \frac{x}{n} = t g i v e$ $A_{n} = \int_{0}^{\infty} e^{- {n n}^{2} t^{2}} c o s (t) n d t$ $= n \int_{0}^{\infty} e^{- n^{3} t^{2}} c o s t d t$ $= \frac{n}{2} \int_{- \infty}^{+ \infty} e^{- n^{3} t^{2} + i t} d t$ $\frac{2}{n} A_{n} = \int_{- \infty}^{+ \infty} e^{- {{(n \sqrt{n} t)}^{2} - i t}} d t$ $= \int_{- \infty}^{+ \infty} e^{- {{(n \sqrt{n} t)}^{2} - 2 \frac{i}{n \sqrt{n}} (n \sqrt{n} t) + {(\frac{i}{n \sqrt{n}})}^{2} - {(\frac{i}{n \sqrt{n}})}^{2}}} d t$ $= \int_{- \infty}^{+ \infty} e^{- {(n \sqrt{n} t - \frac{i}{n \sqrt{n}})}^{2} - \frac{1}{n^{3}}} d t$ $=_{n \sqrt{n} t - \frac{i}{n \sqrt{n}} = u} e^{- \frac{1}{n^{3}}} \int_{- \infty}^{+ \infty} e^{- u^{2}} d u$ $= \sqrt{π} e^{- \frac{1}{n^{3}}} \Rightarrow$ $A_{n} = \frac{n \sqrt{π}}{2} e^{- \frac{1}{n^{3}}} .$
Commented by math khazana by abdo last updated on 17/Jun/18

$2) i t s c l e a r t h a t {l i m}_{n \to + \infty} A_{n} = + \infty .$
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